There are 1,200 students at roosevelt middle school. There are 700 boys at the school. What percentage of the students at roosevelt are boys( to the nearest tenth of a percent)?

Answer:  The answer is:  "  [tex]\frac{7}{80}[/tex] " .

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Step-by-step explanation:

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Given:  " 8 [tex]\frac{3}{4}[/tex] % "  ;  Convert this to a "fraction" .

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Start with the:  " 8  [tex]\frac{3}{4}[/tex] "  portion:

Start with the:  " [tex]\frac{3}{4}[/tex] "  portion :

Note:  " [tex]\frac{3}{4}[/tex] " ;

                           =  3/4 ;

                           =  3 ÷ 4 ;  {use calculator} ;

                           =  0.75  .

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Or;  recognize, from experience, that:  "3/4" = " 0.75 " .

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Or:  Calculate as follows:

          →  " [tex]\frac{3}{4}[/tex]  = [tex]\frac{?}{100}[/tex] " ;  

          →   Solve for the:  "(?)" value:

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Consider the "denominator" portions:

         →  " 4 * {what value?} = 100 " ;

         →  100/4 = 100 ÷ 4  = 25 ;  

         →  So:  " 4 * {25}  = 100 ;

Now, consider the "numerator" portions:

         →  " 3 * {25}  = "{?}" ;

         →  " 3 * {25}  = " 75 " ;

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Now, we can rewrite the expression;

           & replace the "{?}" — with:  " 75 " ;  as follows:

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The original expression:

          →  " [tex]\frac{3}{4}[/tex]  = [tex]\frac{?}{100}[/tex] " ;  

Rewrite as:

          →  " [tex]\frac{3}{4}[/tex]  = [tex]\frac{75}{100}[/tex] " ;  

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Note:  " [tex]\frac{75}{100}[/tex] "  ;

                       =   75 / 100  =  75 ÷ 100 ;  

          →   Use calculator:   = 0.75 ;

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                               Or:  75 ÷ 100 = ? ;

     →  Note:  when dividing by "100" ; we take the original value, and move that number backward "2 (two) decimal spaces".  We move "backward" because we are "dividing" (not "multiplying") ; and we move 2 (two) decimal spaces because:  " 100" has 2 (two) zeros.

     →  " 75 ÷ 100  =  75.  ÷  100  =  .75  ;

                               →  Write as:  " 0.75.

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So:  " 8 [tex]\frac{3}{4}[/tex]  "  ;

          =  8  +  [tex]\frac{3}{4}[/tex]  ;

          =   8  + 0.75 ;

         =   8.75  .

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So:  " 8 [tex]\frac{3}{4}[/tex] % " ;  

          =  8.75 % ;

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Note that " % " ;  refers to parts per hundred;  or:  "divided by 100" ;

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       →  So:   " 8.75 % " ;

                  = 8.75 / 100 = 8.75 ÷ 100 ;  

       →  Use calculator; to get:  " 0.0875 " .

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Or:  " 8.75 ÷ 100 " ;   →  As aforementioned, when dividing by "100" ;

                                   we move the decimal point "backward 2 (two) spaces:

      →   " 8.75 ÷ 100  =  .0875 ;

                          →  Write as:  " 0.0875 " .

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Now, let us convert this value to a "fraction value" ;

    →  We have:  " 0.0875 ;

We look at the last consecutive non-zero digits:  "875" .

   →  How many decimal spaces backward from "875" ;  specifically, the digit, "5" ;  are there (counting backward; to the actual decimal point)?

  →  0.0875 .  →  The answer is:  "4 (four)" ;  which are: 5, 7, 8, and 0 " .

                               {Refer to digits in "bold font".}.

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So, this value:  " 0.0875 " ;  

          →  is equal to:  " 875 " ÷ { 1 followed by 4 [four] zeros } ;

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                               =  " 875 ÷ 10000 " ;

                               = " 875 ÷ 10,000 " ;

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  Using a calculator:

       " 875 ÷ 10,000 "  = " [tex]\frac{875}{10,000}[/tex] " ;

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                                    = " [tex]\frac{7}{80}[/tex] " .

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The answer is: "  [tex]\frac{7}{80}[/tex] " .

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Hope this is helpful to you!

       Best wishes to you!

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8 3/4% as a fraction in simplest form is 7/80.

In Mathematics and Geometry, a fraction simply refers to a numerical quantity (numeral) which is not expressed as a whole number. This ultimately implies that, a fraction is simply a part of a whole number.

In Mathematics, a percentage is any number that is expressed as a fraction of hundred (100). This ultimately implies that, a percentage indicates the hundredth parts of any given number.

In this exercise and scenario, we would convert the given percentage 8 3/4% into a fraction by dividing as follows;

8 3/4% = 8% + 0.75%

8% + 0.75% = 8.75%

8.75/100

Multiply both sides by 100, we have:

Fraction: 875/10000 = 7/80.

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Base Case: plug in n = 1 (the smallest positive integer)

If n = 1, then 3n-2 = 3*1-2 = 1. Square this and we see that (3n-2)^2 = 1^2 = 1

On the right hand side, plugging in n = 1 leads to...

n*(6n^2-3n-1)/2 = 1*(6*1^2-3*1-1)/2 = 1

Both sides are 1. So that confirms the base case.

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Inductive Step: Assume that

1^2 + 4^2 + 7^2 + ... + (3k-2)^2 = k*(6k^2-3k-1)/2

is a true statement for some positive integer k. If we can show the statement leads to the (k+1)th case being true as well, then we will have sufficiently proven the overall statement to be true by induction.

1^2 + 4^2 + 7^2 + ... + (3k-2)^2 = k*(6k^2-3k-1)/2

1^2 + 4^2 + 7^2 + ... + (3k-2)^2 + (3(k+1)-2)^2 = (k+1)*(6(k+1)^2-3(k+1)-1)/2

k*(6k^2-3k-1)/2 + (3(k+1)-2)^2 = (k+1)*(6(k^2+2k+1)-3(k+1)-1)/2

k*(6k^2-3k-1)/2 + (3k+3-2)^2 = (k+1)*(6k^2+12k+6-3k-3-1)/2

k*(6k^2-3k-1)/2 + (3k+1)^2 = (k+1)*(6k^2+9k+2)/2

k*(6k^2-3k-1)/2 + 9k^2+6k+1 = (k+1)*(6k^2+9k+2)/2

(6k^3-3k^2-k)/2 + 2(9k^2+6k+1)/2 = (k*(6k^2+9k+2)+1(6k^2+9k+2))/2

(6k^3-3k^2-k + 2(9k^2+6k+1))/2 = (6k^3+9k^2+2k+6k^2+9k+2)/2

(6k^3-3k^2-k + 18k^2+12k+2)/2 = (6k^3+9k^2+2k+6k^2+9k+2)/2

(6k^3+15k^2+11k+2)/2 = (6k^3+15k^2+11k+2)/2

Both sides simplify to the same expression, so that proves the (k+1)th case immediately follows from the kth case

That wraps up the inductive step. The full induction proof is done at this point.

Answer:

D. 7

Step-by-step explanation:

In synthetic division, the first # in the division bracket (2)  is dropped, and the number outside the bracket is multiplied with it to find 4. Then, -2 and 4 are added to get 2. Again, 2 x 2=4. 3+4=7.

Simplified version: Drop, multiply, add, multiply, add till you reach the last term.

The synthetic division problem hasn't been stated in this query, making it impossible to ascertain the missing number. Synthetic division is a mathematical methodology used to simplify polynomial division by binomial. The correct problem should be offered for accurate guidance.

Unfortunately, the provided synthetic division problem was not included in your question so I'm unable to solve for the missing number. In general, synthetic division is a method used in algebra to divide a polynomial by a binomial. I would suggest you to verify the problem and restate your question providing all necessary details. Then I would gladly help you solve the missing number in the synthetic division.

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The inequality to represent the situation is 335 + d ≥ 575; Rhonda must collect at least '240 stamps' in order to reach her goal.

If the equation or inequality contains variable terms, then there might be some values of those variables for which that equation or inequality might be true. Such values are called solution to that equation or inequality. Set of such values is called solution set to the considered equation or inequality.

Given that the Stamps Rhonda wants to collect = 575

The Stamps already in collection = 335

Let consider 'd' represents the number of stamps Rhonda needs to achieve her goal.

Thus, the Stamps to collect + Stamps already in collection ≥ 575

d + 335 ≥ 575

d ≥ 575 - 335

d ≥ 240

Hence, The inequality to represent the situation is 335 + d ≥ 575; Rhonda must collect at least '240 stamps' in order to reach her goal.

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Final answer:

To find how many more stamps Rhonda needs to collect to reach her goal, utilize the inequality 335 + d ≥ 575 with d representing the additional stamps needed, resulting in d ≥ 240.

Explanation:

Rhonda's Inequality:

335 + d ≥ 575; d ≥ 240

To determine how many more stamps Rhonda must collect to reach her goal, we set up the inequality 335 + d ≥ 575, where d represents the number of stamps she must collect. By solving this inequality, we find that Rhonda needs to collect at least 240 more stamps to reach her goal.

Answer:

5 pages

Step-by-step explanation:

30 sticker and 6 stickers on each page

30 divided by 6 = 5

Answer:

He puts stickers on 5 pages.

Answer:

18

Step-by-step explanation:

Since this is an isosceles triangle, sides AB and sides AC are equal

AB=AC

8x-4 = 5x+11

Subtract 5x from each side

8x-5x-4 = 5x-5x+11

3x-4 = 11

Add 4 to each side

3x-4+4 = 11+4

3x=15

Divide by 3

3x/3 =15/3

x=5

We want to find BC

BC= 4x-2

Since x=5

BC = 4(5) -2

BC = 20-2

BC =18

Answer:

See proof below

Step-by-step explanation:

Consider triangle with midpoints D, E, F of the sides BC, AC and AB, respectively.  If D, E and F are midpoints  of the sides BC, AC and AB, then

Triangle ABC is equilateral triangle, then

If AB=BC=AC, then EA=CE=FA=BF=DC=DB.

By SAS theorem, ΔFAE≅ΔDCE≅ΔEBD.

Congruent triangles have congruent corresponding sides, then

EF=FD=DE. This means that triangle DEF is equilateral.

To prove that △RST ≅ △UVW by ASA (Angle-Side-Angle) congruence, we need to show that angle S is congruent to angle U, side ST is congruent to side VW, and angle R is congruent to angle V.

To prove that △RST ≅ △UVW by ASA (Angle-Side-Angle) congruence, we need to show that angle S is congruent to angle U, side ST is congruent to side VW, and angle R is congruent to angle V.

1. Given that ∠R ≅ ∠U, we have the first angle in both triangles congruent.

2. Given that ST ≅ VW, we have the side opposite to the congruent angles congruent in both triangles.

3. Given that ∠S ≅ ∠V, we have the second angle in both triangles congruent.

Using ASA congruence, we can conclude that △RST ≅ △UVW.

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Final answer:

Triangles △RST and △UVW are congruent by ASA postulate, which states if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the two triangles are congruent.

Explanation:

To prove that triangles △RST and △UVW are congruent by the Angle-Side-Angle (ASA) postulate, we first observe the given congruent parts: ∠R ≅ ∠U, ST ≅ VW, and ∠S ≅ ∠V. Now, by the ASA postulate, if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the two triangles are congruent. We have two pairs of congruent angles: ∠R ≅ ∠U and ∠S ≅ ∠V, and one pair of congruent sides: ST ≅ VW (△RST's side ST is included between ∠R and ∠S, and △UVW's side VW is included between ∠U and ∠V). This satisfies the requirements for the ASA postulate, so we can conclude the triangles are congruent: △RST ≅ △UVW.

Answer:

x = 6

Step-by-step explanation:

AD and EH are equal. Congruent figures are made up of congruent parts.

Let the initial savings at first be = x

Given ratio is [tex]\frac{5}{4}[/tex] and after donating $40 to charity, the ratio becomes [tex]\frac{13}{10}[/tex]

Hence, the relation becomes:

[tex]\frac{5x-40}{4x-40}=\frac{13}{10}[/tex]

[tex]10(5x-40)=13(4x-40)[/tex]

= [tex]50x-400=52x-520[/tex]

[tex]-2x=-120[/tex] or

[tex]2x=120[/tex]

[tex]x=60[/tex]

So , Dave's saving at first was 5x = 5*60 = $300

and Sam's saving at first was 4x = 4*60= $240

Answer: there is an app that helps you with this stuff. but are you solving for x or y

Step-by-step explanation:

Answer:

1/6 = 3/18

1/9 = 2/18

Step-by-step explanation:

The common denominator for 6 and 9 is the smallest number that they both go into, which is 18.  We need to multiply 6 by 3 to get 18 and multiply 9 by 2 to get 18

1/6 *3/3 = 3/18

1/9 *2/2 = 2/18

Answer:

the pair would be 3/18 and 2/18

Divide 260 and 156 to get 5/3, or 1 2/3. Multiply by 100 to get 166.(6). (6) means repeating forever. For example 1.(3)=1.333333333....

Answer:

2 lbs  9 oz

Step-by-step explanation:

Lets line up the ounces and pounds and subtract

  5 lb. 10 oz.

− 3 lb. 1 oz.

------------------------

2 lbs  9 oz

The result of subtraction of 5 lb. 10 oz. − 3 lb. 1 oz.  is: 2 lbs  9 oz.

Here, we have,

given that,

5 lb. 10 oz. − 3 lb. 1 oz.

so, we have to subtract them.

Lets line up the ounces and pounds and subtract

 5 lb. 10 oz.

− 3 lb. 1 oz.

------------------------

2 lbs  9 oz

Hence, The result of subtraction of 5 lb. 10 oz. − 3 lb. 1 oz.  is: 2 lbs  9 oz.

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Answer:

$780

Step-by-step explanation:

The initial sum of money borrowed was $750

at a simple interest rate of 8%

Dan pays the money back in 6 months

The amount of money Dan has to pay back his brother including the interest in 6 months' time.

The total amount accrued, principal plus interest, from simple interest on a principal of $750 at a rate of 8% per year for 6 months is $780.

The total interest that is to be paid back can be calculated using the formula

where P is the prinicipal amount of money we begin with (in this case $750), r is the rate of interest (in this case 8%) and t is the time period (in this case, 6 months or 1/2 of a year).

(Note that we must take the value of t in years if the interest is annual interest and that is why we use t = 1/2 = 0.5 instead of t = 6).

So, by simple calculation we can find interest to be paid as

the interest accrued is $30

To find the total sum to be paid back, we add the initial sum or pricipal amount to the interest we have calculated above i.e.,

the total amount accrued, principal plus interest, from simple interest on a principal of $750 at a rate of 8% per year for 6 months is $780.

Answer:

b=2A-2a

Step-by-step explanation:

First you subtract a to the other side.

Then you divide by 2.

SO it will be b=2A-2a which is your answer

Answer:

The days will it take them to have the same number of followers is:

Step-by-step explanation:

To solve the proposed exercise, an equation must be made for each one taking into account the information obtained:

Where d is the number of days elapsed. Since you want to identify the day in which they will have the same number of followers, you proceed to match the two equations:

And you clear the variable d (remember that those who are adding to one side of equality, go to the other side to subtract and, if they are subtracting to one side of equality, go to add):

Answer:

D - it uses data entered in certain cells to calculate values for other cells

Step-by-step explanation:

A-p-e-x approved

Answer:

c. 1/4 or 25% or 0.25

Step-by-step explanation:

Answer:

C  1/4

Step-by-step explanation:

When the bases are the same, we can add the exponents

x^a * x^b = x^ (a+b)

4^2 * 4^(-3)

4^(2-3)

4^(-1)

The negative exponent puts it in the denominator

1/4

Answer:

Third choice "The ratio given is part to part. The total number of parts in the whole is 1 + 3 = 4." is the correct answer.

Step-by-step explanation:

We know that partitioning a directed line segment into a ratio of 1:3 means that we are dividing the given line segment into two parts whose first part is 1 times the of some quantity while the another part is 3 times of the same quantity. So basically we are comparing part to part in by ratio. And total number of parts in the whole will be just sum of both so we get 1+3=4

Hence choice (3) is correct answer.

"The ratio given is part to part. The total number of parts in the whole is 1 + 3 = 4."

Answer:

it is C

The ratio given is part to part. The total number of parts in the whole is 1 + 3 = 4." is the correct answer.

Final answer:

The jogger runs a total distance of 224 meters, which includes running along the width and length of the park and taking a shortcut diagonally across the park.

Explanation:

To calculate the total distance the jogger runs, we must add the distances from each leg of the jogger's path. Starting from the southeast corner, the jogger runs the width of the park (28 meters) to the west, and then the length of the park (96 meters) to the north, reaching the northwest corner. The shortcut from the northwest corner directly back to the southeast corner, where her car is parked, is the hypotenuse of the resulting right-angled triangle. To find the length of this hypotenuse, we use the Pythagorean theorem (a² + b² = c²), where the sides of the rectangle serve as 'a' and 'b'.

Let's calculate the hypotenuse:

a = 28 m (width)

b = 96 m (length)

c = √(a² + b²) = √(28² + 96²)

c = √(784 + 9216)

c = √10000

c = 100 m

Now, we sum up all the distances:

28 m (west along the width)

96 m (north along the length)

100 m (shortcut diagonally)

Total distance = 28 m + 96 m + 100 m = 224 meters

Answer:

Cost of 11 cans of Dog food is $9.57

Step-by-step explanation:

Cost of 5 cans of dog food = $4.35

So to find out the cost of 11 cans of dog food at the same price, we need to calculate the price of 1 can of dog food.

Calculation for finding out cost of 1 can of dog food

The price $4.35 which is given is the price of 5 cans

So to find the cost of 1 can, we need to divide the given price by 5

5 Cans of dog food = $4.35

1 can of dog food = 4.35 ÷ 5

1 can of dog food = $0.87

Now to find out the price of 11 cans, we will multiply the price of 1 can with 11

Cost of 11 cans = 11 × $0.87

                          = $9.57

Answer:

The null and alternative hypotheses are:

[tex]H_{0}: \mu = 170[/tex]

[tex]H_{a}: \mu >170[/tex]

Under the null hypothesis, the test statistic is:

[tex]z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

         [tex]=\frac{180-170}{\frac{35}{\sqrt{225}} }[/tex]

         [tex]=\frac{10}{2.33}[/tex]

         [tex]=4.29[/tex]

Now, we have to find the right tailed z critical value at 0.10 significance level. Using the standard normal table, we have:

[tex]z_{critical} = 1.28[/tex]

Since the test statistic is greater than the z critical value, we therefore, reject the null hypothesis and conclude that there is sufficient evidence to support the claim that the university students weigh more than the population.

Answer:

Correct choice is D

Step-by-step explanation:

Consider the quadratic expression [tex]5x^2 + 7x + 2.[/tex]

First, find the discriminant:

[tex]D=7^2-4\cdot 5\cdot 2=49-40=9.[/tex]

Now find the roots of the quadratic expression [tex]5x^2 + 7x + 2:[/tex]

[tex]x_{1,2}=\dfrac{-7\pm \sqrt{9}}{2\cdot 5}=\dfrac{-7\pm 3}{10}=-1,\ -0.4.[/tex]

Then

[tex]5x^2 + 7x + 2=5(x-(-1))(x-(-0.4))=5(x+1)(x+0.4)=(x+1)(5x+2).[/tex]

Answer:

Proof

Step-by-step explanation:

Recall the identity [tex]\sin(a \pm b) = \sin(a)\cos(b) \pm \cos(a)\sin(b)[/tex].

Consider [tex]\sin(a + b) = k\sin(a - b)[/tex]. We firstly apply the above identity to reach

[tex]\sin(a)\cos(b) + \cos(a)\sin(b) = k(\sin(a)\cos(b) - \cos(a)\sin(b))[/tex].

By expanding the bracket on the right we obtain

[tex]\sin(a)\cos(b) + \cos(a)\sin(b) = k\sin(a)\cos(b) - k\cos(a)\sin(b)[/tex] and so

[tex]\cos(a)\sin(b) + k\cos(a)\sin(b) = k\sin(a)\cos(b) - \sin(a)\cos(b)[/tex] and so

[tex](1+k)\cos(a)\sin(b) = (k-1)\sin(a)\cos(b)[/tex] and so

[tex](k+1)\frac{\cos(a)}{\sin(a)}= (k-1)\frac{\cos(b)}{\sin(b)}[/tex] and finally

[tex](k+1)\cot(a)= (k-1)\cot(b)[/tex].

Final answer:

Using the sum and difference identities for sine, we can manipulate the equation sin(a+b) = k sin(a-b) to prove the trigonometric identity (k-1)cotb = (k+1)cota as required.

Explanation:

To solve the given trigonometric identity, we must show that if sin(a+b) = k sin(a-b), then (k-1)cotb = (k+1)cota. First, let's utilize the sum and difference identities for sine:

sin(a+b) = sin(a)cos(b) + cos(a)sin(b)

sin(a-b) = sin(a)cos(b) - cos(a)sin(b)

Substitute these into the original equation:

sin(a)cos(b) + cos(a)sin(b) = k(sin(a)cos(b) - cos(a)sin(b))

Now, distribute the k on the right side and arrange terms:

sin(a)cos(b)(1-k) = cos(a)sin(b)(1+k)

Divide both sides by sin(b)cos(b) to isolate cot(a) and cot(b):

(1-k) cot(b) = (1+k) cot(a)

This proves the identity as required. Notice that the cotangent function is the reciprocal of the tangent function, which is the ratio of sine to cosine.

The type of angle that is not formed by two parallel lines cut by a transversal is complementary angles. So option (d) is correct.

The intersection of two parallel lines by a transversal forms several types of angles:

1. Alternate Interior Angles: These are angles on opposite sides of the transversal and inside the parallel lines. They are congruent.

2. Alternate Exterior Angles: These are angles on opposite sides of the transversal and outside the parallel lines. They are also congruent.

3. Corresponding Angles: These are angles in the same position at each intersection. They are congruent when the parallel lines are cut by a transversal.

4. Complementary Angles: These are angles whose measures sum up to 90 degrees. They can be formed when a transversal intersects parallel lines, but they are not a specific type of angle formed by this intersection.

Therefore, the type of angle that is not formed by two parallel lines cut by a transversal is D) complementary angles. The other options, A) alternate interior angles, B) alternate exterior angles, and C) corresponding angles, are all formed by this intersection.

Answer: 12

Explanation: what i did to get that was i did 60 divided by 5 = 12 or u can do 12 times 5 which will give you 60 mins

hope this helps you Good Luck!!!!!!!!